# Volume of a sphere using cartesian coordinates

i'm preparing my calculus exam and i'm in doubt about how to generally compute triple integrals. I know that the cartesian equation of a sphere is $$B_R=\{(x, y, z)|x^2+y^2+z^2=R^2\}$$, so (if i didn't want to use spherical coordinates, wich i'm aware is the best way and i already did that) its volume would just be $$\iiint_S dxdydz$$, but what would the extremes be? I know $$-R \leq z \leq R$$ and $$-\sqrt{R^2-y^2-z^2} \leq x \leq\sqrt{R^2-y^2-z^2}$$, but what are the extremes for $$y$$? I can't describe it in terms of $$x$$, so i have $$Vol(B_R) = \int_{-R}^{R} \int_?^?\int_{-\sqrt{R^2-y^2-z^2}}^\sqrt{R^2-y^2-z^2} dxdydz.$$ What should be there instead of the '?'

• The limits are $\pm\sqrt{R^2-z^2}$, because that is the maximum/minimum that $y$ is allowed to take knowing $z$ only. – user10354138 Jun 8 at 15:51
• So i have to compute the maximum/minimum of the function and substitute its $x$ coordinate in the extremes? – DottorMaelstrom Jun 8 at 15:55
• I don't know what "function" you are talking about. Also, sphere is 2-dimensional, so doesn't have "volume". You want the ball $B_R=\{x^2+y^2+z^2\leq R^2\}$ instead. – user10354138 Jun 8 at 16:01

After doing away with the integral over $$x$$ it remains to be integrated over $$y$$ and $$z$$, as you already know. The domain of the remaining double integral is a circle on the $$YZ$$ plane. So, its limits are $$-\sqrt{R^2 - z^2}$$ (lower) and $$\sqrt{R^2 - z^2}$$ (upper).

If you want to see it geometrically, think of your 2-dimensional sphere of radius $$R$$ in the 3-dimensional Cartesian space. Project everything onto the $$YZ$$ plane to get a flat 2-dimensional circle of radius $$R$$. You can see that circle as the union of infinitely many half-circumferences of radius $$R$$, for $$R$$ ranging from $$0$$ to $$R$$.

• So in general when computing multiple integrals, after being done with one variable, i should "forget it" and think the space as losing one dimension (the direction of the variable i already integrated)? – DottorMaelstrom Jun 8 at 16:11
• Yes, but let me elaborate a bit on that. By integrating on a variable (say $x$) with those limits, you are passing the domain's information onto the integrand. When integrating a definite integral over $x$ your result cannot depend on $x$. With iterated integrals we follow this process with the hopes of obtaining a real number, which is the area or volume of a geometric object. – Sam Skywalker Jun 8 at 16:21

It should be $$$$\text{vol}(B_R) = \int_{-R}^{R} \int_{-\sqrt{R^2-z^2}}^{\sqrt{R^2-z^2}} \int_{-\sqrt{R^2-z^2-y^2}}^{\sqrt{R^2-z^2-y^2}} \, dx \, dy \, dz$$$$ The way to think about this is to successively "fix" each variable as follows: pick a variable, for example $$z$$. Clearly, its bounds are $$-R \leq z \leq R$$. Then, for a fixed $$z$$, we have $$$$0 \leq x^2 + y^2 \leq R^2 - z^2.$$$$ Now, pick your next variable, say $$y$$. From the above inequality, it follows that $$y$$ must satisfy $$- \sqrt{R^2 - z^2} \leq y \leq \sqrt{R^2 - z^2}$$ (if it was outside this interval, the inequality above would be violated). Now, keep $$y$$ fixed. Then, we get $$$$0 \leq x^2 \leq R^2 - z^2 - y^2$$$$ which implies $$-\sqrt{R^2 - z^2 - y^2} \leq x \leq \sqrt{R^2 - z^2 - y^2}$$.

So, what you're doing is you look at the constraint all $$3$$ variables satisfy. Then, you choose one of them (for example $$z$$) and ask yourself what are the bounds it can take (in this case it was obvious). Then, you think of $$z$$ as fixed, and ask yourself what constraint do the remaining two variables satisfy (in this case $$x,y$$). Then you pick one of them (for example $$y$$) and ask what values can it take. Now, think of $$y$$ as fixed ax well, so that you end up with a constraint on $$x$$. This is a step-by-step reduction process which should be helpful; to really solidify your understanding you should see what each step is doing geometrically as well.